# How do you solve using gaussian elimination or gauss-jordan elimination, 2x–3y+2z=2, x+4y-z=9, -3x+y–5z=5?

Feb 14, 2018

$x = \frac{115}{31}$, $y = \frac{32}{31}$ and $z = - \frac{36}{31}$

#### Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

$A = \left(\begin{matrix}2 & - 3 & 2 & | & 2 \\ 1 & 4 & - 1 & | & 9 \\ - 3 & 1 & - 5 & | & 5\end{matrix}\right)$

I have written the equations not in the sequence as in the question in order to get $1$ as pivot.

Perform the folowing operations on the rows of the matrix

$R 1 \leftarrow R 1 - 2 R 2$; $R 3 \leftarrow R 3 + 3 R 2$

$A = \left(\begin{matrix}0 & - 11 & 4 & | & - 16 \\ 1 & 4 & - 1 & | & 9 \\ 0 & 13 & - 16 & | & 32\end{matrix}\right)$

$R 3 \leftarrow \left(R 3 + R 1\right)$

$A = \left(\begin{matrix}0 & - 11 & 4 & | & - 16 \\ 1 & 4 & - 1 & | & 9 \\ 0 & 2 & - 12 & | & 16\end{matrix}\right)$

$R 3 \leftarrow \frac{R 3}{2}$

$A = \left(\begin{matrix}0 & - 11 & 4 & | & - 16 \\ 1 & 4 & - 1 & | & 9 \\ 0 & 1 & - 6 & | & 8\end{matrix}\right)$

R1larrR1+11R3; R2larrR2-4R3

$A = \left(\begin{matrix}0 & 0 & - 62 & | & 72 \\ 1 & 0 & 23 & | & - 23 \\ 0 & 1 & - 6 & | & 8\end{matrix}\right)$

$R 1 \leftarrow \frac{R 1}{- 62}$

$A = \left(\begin{matrix}0 & 0 & 1 & | & - \frac{36}{31} \\ 1 & 0 & 23 & | & - 23 \\ 0 & 1 & - 6 & | & 8\end{matrix}\right)$

R2larrR2-23R1; R3larrR3+6R1

$A = \left(\begin{matrix}0 & 0 & 1 & | & - \frac{36}{31} \\ 1 & 0 & 0 & | & \frac{115}{31} \\ 0 & 1 & 0 & | & \frac{32}{31}\end{matrix}\right)$

Thus, $x = \frac{115}{31}$, $y = \frac{32}{31}$ and $z = - \frac{36}{31}$